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# Finding Zeros of a Polynomial Function

### Learning Outcomes

• Use synthetic division to find the zeros of a polynomial function.
• Use the Fundamental Theorem of Algebra to find complex zeros of a polynomial function.
The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function.

### How To: Given a polynomial function $f$, use synthetic division to find its zeros

1. Use the Rational Zero Theorem to list all possible rational zeros of the function.
2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.
3. Repeat step two using the quotient found from synthetic division. If possible, continue until the quotient is a quadratic.
4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.

### Example: Finding the Zeros of a Polynomial Function with Repeated Real Zeros

Find the zeros of $f\left(x\right)=4{x}^{3}-3x - 1$.

Answer: The Rational Zero Theorem tells us that if $\frac{p}{q}$ is a zero of $f\left(x\right)$, then is a factor of –1 and q is a factor of 4.

$\begin{array}{l}\frac{p}{q}=\frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}}\hfill \\ \text{}\frac{p}{q}=\frac{\text{Factors of -1}}{\text{Factors of 4}}\hfill \end{array}$

The factors of –1 are $\pm 1$ and the factors of 4 are $\pm 1,\pm 2$, and $\pm 4$. The possible values for $\frac{p}{q}$ are $\pm 1,\pm \frac{1}{2}$, and $\pm \frac{1}{4}$. These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1. Dividing by $\left(x - 1\right)$ gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as $\left(x - 1\right)\left(4{x}^{2}+4x+1\right)$. The quadratic is a perfect square. $f\left(x\right)$ can be written as $\left(x - 1\right){\left(2x+1\right)}^{2}$. We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.

$\begin{array}{l}2x+1=0\hfill \\ \text{ }x=-\frac{1}{2}\hfill \end{array}$

The zeros of the function are 1 and $-\frac{1}{2}$ with multiplicity 2.

#### Analysis of the Solution

Look at the graph of the function f. Notice, at $x=-0.5$, the graph bounces off the x-axis, indicating the even multiplicity (2,4,6…) for the zero –0.5. At $x=1$, the graph crosses the x-axis, indicating the odd multiplicity (1,3,5…) for the zero $x=1$.

## The Fundamental Theorem of Algebra

Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations. Suppose f is a polynomial function of degree four and $f\left(x\right)=0$. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it ${c}_{1}$. By the Factor Theorem, we can write $f\left(x\right)$ as a product of $x-{c}_{\text{1}}$ and a polynomial quotient. Since $x-{c}_{\text{1}}$ is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it ${c}_{\text{2}}$. We can write the polynomial quotient as a product of $x-{c}_{\text{2}}$ and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of $f\left(x\right)$.

### A General Note: The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that, if $f(x)$ is a polynomial of degree $n>0$, then $f(x)$ has at least one complex zero. We can use this theorem to argue that, if $f\left(x\right)$ is a polynomial of degree $n>0$, and a is a non-zero real number, then $f\left(x\right)$ has exactly n linear factors. The polynomial can be written as

$f\left(x\right)=a\left(x-{c}_{1}\right)\left(x-{c}_{2}\right)...\left(x-{c}_{n}\right)$

where ${c}_{1},{c}_{2},...,{c}_{n}$ are complex numbers. Therefore, $f\left(x\right)$ has n roots if we allow for multiplicities.

### recall complex numbers

Recall that we defined complex numbers as numbers of the form $a + bi$. To graph such numbers required the complex plane, made up of a real axis and an imaginary axis. This plane defined the real numbers as a subset of the complex numbers, just as the rational and irrational numbers are subsets of the real numbers. In other words, each real number is also a complex number of the form $a+bi$, where $b=0$.

### Q & A

Does every polynomial have at least one imaginary zero? No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.

### Example: Finding the Zeros of a Polynomial Function with Complex Zeros

Find the zeros of $f\left(x\right)=3{x}^{3}+9{x}^{2}+x+3$.

Answer: The Rational Zero Theorem tells us that if $\frac{p}{q}$ is a zero of $f\left(x\right)$, then p is a factor of 3 and q is a factor of 3.

$\begin{array}{l}\frac{p}{q}=\frac{\text{Factors of the constant term}}{\text{Factor of the leading coefficient}}\hfill \\ \text{}\frac{p}{q}=\frac{\text{Factors of 3}}{\text{Factors of 3}}\hfill \end{array}$

The factors of 3 are $\pm 1$ and $\pm 3$. The possible values for $\frac{p}{q}$, and therefore the possible rational zeros for the function, are $\pm 3, \pm 1, \text{and} \pm \frac{1}{3}$. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with –3. Dividing by $\left(x+3\right)$ gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as $\left(x+3\right)\left(3{x}^{2}+1\right)$. We can then set the quadratic equal to 0 and solve to find the other zeros of the function.

$\begin{array}{l}3{x}^{2}+1=0\hfill \\ \text{ }{x}^{2}=-\frac{1}{3}\hfill \\ \text{ }x=\pm \sqrt{-\frac{1}{3}}=\pm \frac{i\sqrt{3}}{3}\hfill \end{array}$

The zeros of $f\left(x\right)$ are –3 and $\pm \frac{i\sqrt{3}}{3}$.

#### Analysis of the Solution

Look at the graph of the function f. Notice that, at $x=-3$, the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero $x=-3$. Also note the presence of the two turning points. This means that, since there is a 3rd degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of $x=-3$ is 1 and there are two complex solutions, which is what we found, or the multiplicity at $x=-3$ is three. Either way, our result is correct.

### Try It

Find the zeros of $f\left(x\right)=2{x}^{3}+5{x}^{2}-11x+4$.

Answer: The zeros are $\text{-4, }\frac{1}{2},\text{ and 1}\text{.}$

[ohm_question]103644[/ohm_question]